Упр.5.491 ГДЗ Виленкин Жохов 5 класс Часть 2, Просвещение (Математика)

Задача

Найдите значение выражения: а) $$\frac{61}{64}-\left(\frac{7}{12}-\frac{5}{14}\right)\cdot\left(\frac{13}{16}+\frac{1}{2}\right)$$; в) $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}$$; б) $$\left(1-\frac{11}{17}\right)\cdot\left(\frac{3}{4}-\frac{5}{12}+\frac{11}{18}\right)$$; г) $$\frac{1}{8}+\frac{3}{8}+\frac{1}{12}+\frac{5}{12}+\frac{1}{16}+\frac{7}{16}+\frac{1}{20}+\frac{9}{20}$$.

Подробный ответ

а) Преобразуем скобки:

$$\frac{7}{12}-\frac{5}{14}=\frac{49}{84}-\frac{30}{84}=\frac{19}{84}, \qquad \frac{13}{16}+\frac{1}{2}=\frac{13}{16}+\frac{8}{16}=\frac{21}{16}.$$

Тогда

$$\frac{61}{64}-\left(\frac{7}{12}-\frac{5}{14}\right)\cdot\left(\frac{13}{16}+\frac{1}{2}\right) = \frac{61}{64}-\frac{19}{84}\cdot\frac{21}{16}.$$

$$\frac{19}{84}\cdot\frac{21}{16}=\frac{19\cdot 1}{4\cdot 16}=\frac{19}{64},$$

$$\frac{61}{64}-\frac{19}{64}=\frac{42}{64}=\frac{21}{32}.$$

б)

$$\left(1-\frac{11}{17}\right)\cdot\left(\frac{3}{4}-\frac{5}{12}+\frac{11}{18}\right) = \frac{6}{17}\cdot\left(\frac{27}{36}-\frac{15}{36}+\frac{22}{36}\right).$$

$$\frac{27}{36}-\frac{15}{36}+\frac{22}{36}=\frac{34}{36}.$$

Тогда

$$\frac{6}{17}\cdot\frac{34}{36}=\frac{2}{3}.$$

в)

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6} = \frac{60}{60}-\frac{30}{60}+\frac{20}{60}-\frac{15}{60}+\frac{12}{60}-\frac{10}{60}.$$

$$=\frac{30}{60}+\frac{20}{60}-\frac{15}{60}+\frac{12}{60}-\frac{10}{60} =\frac{50}{60}-\frac{15}{60}+\frac{12}{60}-\frac{10}{60}$$

$$=\frac{35}{60}+\frac{12}{60}-\frac{10}{60} =\frac{47}{60}-\frac{10}{60} =\frac{37}{60}.$$

г)

$$\frac{1}{8}+\frac{3}{8}+\frac{1}{12}+\frac{5}{12}+\frac{1}{16}+\frac{7}{16}+\frac{1}{20}+\frac{9}{20}$$

Сгруппируем слагаемые:

$$\left(\frac{1}{8}+\frac{3}{8}\right)+\left(\frac{1}{12}+\frac{5}{12}\right)+\left(\frac{1}{16}+\frac{7}{16}\right)+\left(\frac{1}{20}+\frac{9}{20}\right).$$

$$=\frac{4}{8}+\frac{6}{12}+\frac{8}{16}+\frac{10}{20} =\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2.$$